Remove the "ine" suffix from the name of the nonmetallic element and add "hydro" as a prefix and "ic" as a suffix.

Representative Oxoacids(ternary acids containing hydrogen, oxygen, and another element, where oxygen and the other element form a common anion such as ClO3- [chlorate], NO3- [nitrate], CO32- [carbonate], SO42- [sulfate], or PO43- [phosphate])

Replace the "ate" suffix in the name of the anion with the suffix "ic".

Nonrepresentative Oxoacids(ternary acids containing hydrogen, oxygen, and another element, where oxygen and the other element form a less common anion such as ClO- [hypochlorite], ClO2- [chlorite], ClO4- [perchlorate], NO2- [nitrite], or SO32- [sulfite])

Identify the corresponding representative oxoacid. If n is the number of oxygens in the representative oxoacid, then in the nonrepresentative oxoacid where the number of oxygens is:
n+1: Add the prefix "per".
n-1: Replace the suffix "ic" with "ous".
n-2: Add the prefix "hypo" and replace the suffix "ic" with "ous".

1These compounds are acids when dissolved in aqueous solution. In the gas phase or as a pure liquid, they are named as nonacidic molecular compounds such as hydrogen bromide (HBr), hydrogen chloride (HCl), hydrogen fluoride (HF), hydrogen iodide (HI), and dihydrogen sulfide (H2S).
2H2S is an exception. In this case, take the name of the nonmetallic element, sulfur, and add "hydro" as a prefix and "ic" as a suffix.
3H2SO4 and H3PO4 are exceptions.
Back to Main Menu

PROBLEM

A compound containing the elements C, H, N, and O is analyzed. When a 1.2359 g sample is burned in excess oxygen, 2.241 g of CO2(g) is formed. The combustion analysis also showed that the sample contained 0.0648 g of H.
(a) Determine the mass, in grams, of C in the 1.2359 g sample of the compound.
(b) When the compound is analyzed for N content only, the mass percent of N is found to be 28.84 percent. Determine the mass, in grams, of N in the original 1.2359 g sample of the compound.
(c) Determine the mass, in grams, of O in the original 1.2359 g sample of the compound.
(d) Determine the empirical formula of the compound.

What is the theoretical yield of CCl4 if 125 g of CS2 is mixed with 435 g of Cl2? What mass of excess reactant remains when the reaction has gone to completion?

SOLUTION

The balanced equation is

CS2(g) + 3Cl2(g) => S2Cl2(l) + CCl4(l)

The molar mass of CS2 is 76.143 g/mole. The molar mass of Cl2 is 70.9054 g/mole. The molar mass of CCl4 is 153.8218 g/mole. Initially, there is 125 g CS2 = 1.642 mole CS2 and 435 g Cl2 = 6.135 mole Cl2. To use up all of the CS2 would require (3)(1.642 mole) = 4.925 mole Cl2. Since there is 6.135 mole Cl2, the limiting reactant is CS2 and there will be excess Cl2 remaining. According to the above balanced equation, each mole of CS2 yields one mole of CCl4. Thus, 1.642 mole CCl4 = 252.5 g CCl4 is produced. The mass of Cl2 remaining is (6.135 mole - 4.925 mole)(70.9054 g/mole) = 85.79 g.

PROBLEM

Suppose that a stable element with atomic number 119, symbol Q, has been discovered.
(a) Write the ground-state electron configuration for Q.
(b) Would Q be a metal or a nonmetal? Explain in terms of electron configuration.
(c) On the basis of periodic trends, would Q have the largest atomic radius in its group or would it have the smallest? Explain in terms of electronic structure.
(d) What would be the most likely charge of the Q ion in stable ionic compounds?
(e) Write a balanced equation that would represent the reaction of Q with water.
(f) Assume that Q reacts to form a carbonate compound. (i) Write the formula for the compound formed between Q and the carbonate ion, CO32-. (ii) Predict whether or not the compound would be soluble in water. Explain your reasoning.

For element Q (Z = 119), we get

Colors identify subshells which belong to the same diagonal. Square brackets indicate periods in the periodic table.

(b) Q has a single valence electron in the 8s subshell and is therefore an alkali metal.

(c) Q would have the largest atomic radius in its group, the alkali metals.

(d) The charge of Q in stable ionic compounds would be +1.

(e) Q would react with water by replacing the hydrogen according to:

2Q + 2H2O => 2QOH + H2

This is a hydrogen displacement redox reaction.

(f) (i) The formula of the compound formed between Q and the carbonate ion would be Q2CO3. (ii) Carbonates containing alkali metal ions are soluble in aqueous solution, so Q2CO3 would be soluble in water.

There is one lone electron pair and three bonding pairs surrounding the central phosphorus atom. The molecule is trigonal pyramidal and therefore non-planar.

PROBLEM

Write the Lewis structure for each of the following and use it and the valence-shell electron-pair repulsion (VSEPR) model to predict the geometry of the molecule:
(a) carbon dioxide (CO2)
(b) carbon tetrachloride (CCl4)

SOLUTION

(a) The carbon has four valence electrons, while each of the two oxygens has six. Thus, we have a total of sixteen valence electrons to account for. The Lewis structure for carbon dioxide is

There is a double bond between the carbon and each of the oxygens. Since the central atom, carbon, has two double bonds and no lone pairs, the VSEPR model predicts that the carbon dioxide molecule is linear, which is consistent with experiment.

(b) The carbon has four valence electrons, while each of the four chlorines has seven. Thus, we have a total of 32 valence electrons to account for. The Lewis structure for carbon tetrachloride is

There is a single bond between the carbon and each of the chlorines. Since the central atom, carbon, has four single bonds and no lone pairs, the VSEPR model predicts that the carbon tetrachloride molecule is tetrahedral.

PROBLEM

For each of the following hybrid atomic orbitals, identify (i) which pure atomic orbitals are used, (ii) the shape of the hybrid atomic orbitals, (iii) the angle between bonded atoms, and (iv) an example of a molecule which exhibits this type of hybrid atomic orbital, without any lone pairs, and its Lewis structure:
(a) sp
(b) sp2(c) sp3(d) sp3d
(e) sp3d2

PROBLEM

(b) On the basis of the Lewis structures drawn, answer the following questions about the particular molecule indicated:
(i) What is the F-C-F bond angle in CF4?
(ii) What is the hybridization of the valence orbitals of P in PF5?
(iii) What is the geometric shape formed by the atoms in SF4?

(b)(i) Carbon tetrafluoride has four electron pairs surrounding the central carbon and no lone pairs. The geometry is tetrahedral. The F-C-F bond angle is 109.5&deg.

(ii) Phosphorus pentafluoride has five electron pairs surrounding the central phosphorus and no lone pairs. The geometry is trigonal bipyramidal. The hybridization that corresponds to this geometry is sp3d.

(iii) Sulfur tetrafluoride has five electron pairs surrounding the central sulfur. One of these is a lone pair. The corresponding geometry is distorted tetrahedral or seesaw.

(c)(i) Structure 1 has four sigma bonds and one pi bond. The double bond consists of one sigma bond and one pi bond.

(ii) In structure 1, the formal charges are all zero. In structure 2, the formal charges are +1 on the phosphorus, -1 on the oxygen, and zero on the fluorines. When a molecule is neutral, like this one, the Lewis structure in which there are no formal charges is preferred, so structure 1 is the better representation.

The total enthalpy change for the above process is

(b) The bond energy of F2(g) is the energy required to break it up into 2F(g), which is 156.9 kJ. The additional electron in F - increases the repulsion between the electron clouds in F and F -, resulting in a weaker bond in F2- than in F2(g).

PROBLEM

Experiments show that it takes 1656 kJ/mole to break all the bonds in methane (CH4) and 4006 kJ/mole to break all the bonds in propane (C3H8). Based on these data, calculate the average bond energy of the C-C bond.

PROBLEM

An unknown gas is one of three possible gases: nitrogen, hydrogen, or oxygen. For each of the three possibilities, describe the result expected when the gas is tested using a glowing splint (a wooden stick with one end that has been ignited and extinguished, but still contains hot, glowing, partially burned wood).

[from 2005 AP Chemistry Free-Response Question 5]

SOLUTION

nitrogen: The splint will completely stop burning and cool.

hydrogen: The gas will ignite.

oxygen: The splint will start burning again with a flame.

PROBLEM

Summarize the following ideal gas laws:
(a) Boyle's law
(b) Charles's law
(c) Gay-Lussac's law
(d) Dalton's law

SOLUTION

(a) Boyle's law, discovered by English scientist Robert Boyle in 1662, states that, for a fixed amount of gas at constant temperature, the product of the pressure P and the volume V is a constant, or PV = constant.

(b) Charles's law, discovered by French chemist Jacques Charles, states that, for a fixed amount of gas at constant pressure, the volume V is proportional to the absolute temperature T or V/T = constant.

(c) Gay-Lussac's law, discovered by French scientist Joseph Gay-Lussac in 1808, states that the ratio between the combining volumes of gases and the volume of the product, if gaseous, can be expressed in small whole numbers.

(d) Dalton's law of partial pressures, discovered by English scientist and school teacher John Dalton in 1801, states that the total gas pressure of a mixture of gases is equal to the sum of the partial pressures of the gases that make up the mixture.

PROBLEM

A compound which has the empirical formula CH2Br has a vapor density of 6.00 g/L at 375 K and 0.983 atm. Using these data, determine the following:
(a) The molar mass of the compound.
(b) The molecular formula of the compound.

(b) The molar mass of CH2Br is 93.9268 g/mole. Therefore, the molecular formula of the compound is C2H4Br2.

PROBLEM

Each of three beakers contains a 0.1 M solution of one of the following solutes: potassium chloride, silver nitrate, or sodium sulfide. The three beakers are labeled randomly as solution 1, solution 2, and solution 3. Mixing solution 1 with solution 2 results in a black precipitate. Mixing solution 2 with solution 3 produces no reaction.
(a) Write the chemical formula of the black precipitate.
(b) Describe the expected results of mixing solution 1 with solution 3.
(c) Identify each of the solutions 1, 2, and 3.

[from 2005 AP Chemistry Free-Response Question 5]

SOLUTION

KCl, AgNO3, and Na2S are all soluble in aqueous solution. The following table shows the solubility of various combinations of the available ions.

Cl-

NO3-

S2-

K+

KClsoluble

KNO3soluble

K2Ssoluble

Ag+

AgClinsoluble

AgNO3soluble

Ag2Sinsoluble

Na+

NaClsoluble

NaNO3soluble

Na2Ssoluble

(a) There are only two possible precipitates, AgCl and Ag2S. AgCl is a white precipitate. Ag2S is a black precipitate. It forms on silver when it is exposed to the hydrogen sulfide of the atmosphere. So the black precipitate is Ag2S.

(c) In order for Ag2S to form when solutions 1 and 2 are mixed, one of the solutions must be AgNO3 and the other must be Na2S. In order for no reaction to occur when solutions 2 and 3 are mixed, one must be KCl and the other must be Na2S. Solution 2 is common to both mixtures, so it is Na2S. Solution 1 is AgNO3 and solution 3 is KCl.

(b) The result of mixing solutions 1 and 3, AgNO3 and KCl, is to produce a white precipitate, AgCl.
Back to Main Menu

The gram-equivalent weight or equivalent of a compound is the weight in grams of the compound which yields or reacts with one mole of hydrogen ions or hydroxyl ions.

Normality is the number of equivalents of solute per liter of solution.

PROBLEM

(a) What is the gram-formula mass of AlCl3?
(b) What is the gram-equivalent mass of AlCl3?

SOLUTION

(a) The gram-formula mass of AlCl3 is the mass, in grams, of one mole of AlCl3, which is 26.98154 g + (3)(35.453 g) = 133.3405 g

(b) The gram-equivalent mass of AlCl3 is the mass, in grams, of AlCl3 which yields or reacts with one mole of hydrogen or hydroxyl ions. Since each of the three Cl- ions in one molecule of AlCl3 can react with one a hydrogen ion, the gram-equivalent mass of AlCl3 is 1/3 of its gram-formula mass or (1/3)(133.3405 g) = 44.44685 g.

SOLUTION

(a) One mole of H2SO4 contains two moles of H+ and is thus equal to two equivalents of H2SO4. Therefore, one equivalent of H2SO4 is 0.5 mole of H2SO4 or (0.5 mole)[(2)(1.008 g/mole) + 32.07 g/mole + (4)(16.00 g/mole)] = 49.04 g.

(b) One mole of H3PO4 contains three moles of H+ and is thus equal to three equivalents of H3PO4. Therefore, one equivalent of H3PO4 is 1/3 mole of H3PO4 or (1/3 mole)[(3)(1.008 g/mole) + 30.97 g/mole + (4)(16.00 g/mole)] = 32.66 g.

(c) One mole of Ca(OH)2 contains two moles of OH- and is thus equal to two equivalents of Ca(OH)2. Therefore, one equivalent of Ca(OH)2 is 0.5 mole of Ca(OH)2 or (0.5 mole)[40.08 g/mole + (2)(16.00 g/mole) + (2)(1.008 g/mole)] = 37.048 g.

(d) One mole of AlCl3 contains three moles of Cl-, which is capable of reacting with three moles of H+, and is thus equal to three equivalents of AlCl3. Therefore, one equivalent of AlCl3 is 1/3 mole of AlCl3 or (1/3 mole)[26.98 g/mole + (3)(35.45 g/mole)] = 44.44 g.

(e) One mole of K2SO4 contains one mole of SO42-, which is capable of reacting with two moles of H+, and is thus equal to two equivalents of K2SO4. Therefore, one equivalent of K2SO4 is 0.5 mole of K2SO4 or (0.5 mole)[(2)(39.10 g/mole) + 32.07 g/mole + (4)(16.00 g/mole)] = 87.14 g.

(f) One mole of NaCl contains one mole of Cl-, which is capable of reacting with one mole of H+, and is thus equal to one equivalent of NaCl. Therefore, one equivalent of NaCl is (1 mole)(22.99 g/mole + 35.45 g/mole) = 58.44 g.

(g) One mole of Na3PO4 contains one mole of PO43-, which is capable of reacting with three moles of H+, and is thus equal to three equivalents of Na3PO4. Therefore, one equivalent of Na3PO4 is 1/3 mole of Na3PO4 or (1/3 mole)[(3)(22.99 g/mole) + 30.97 g/mole + (4)(16.00 g/mole)] = 54.65 g.

(h) One mole of CuSO4 contains one mole of SO42-, which is capable of reacting with two moles of H+, and is thus equal to two equivalents of CuSO4. Therefore, one equivalent of CuSO4 is 0.5 mole of CuSO4 or (0.5 mole)[63.55 g/mole + 32.07 g/mole + (4)(16.00 g/mole)] = 79.81 g.

(i) One mole of KOH contains one mole of OH- and is thus equal to one equivalent KOH. Therefore, one equivalent of KOH is (1 mole)(39.10 g/mole + 16.00 g/mole + 1.008 g/mole) = 56.11 g.

(j) One mole of Fe2O3 contains two moles of Fe3+, which is capable of reacting with six moles of OH-, and is thus equal to six equivalents of Fe2O3. Therefore, one equivalent of Fe2O3 is 1/6 mole of Fe2O3 or (1/6 mole)[(2)(55.85 g/mole) + (3)(16.00 g/mole)] = 26.62 g.

(d) Each mole of Mg(OH)2 contains two moles of OH- ions. Therefore, each mole of Mg(OH)2 is equal to two equivalents of Mg(OH)2. The normality of a 0.4 M Mg(OH)2 solution is 0.8 N.

PROBLEM

If 1 mole of each of the following substances was dissolved in 1000 g of water, what would be the boiling point?
(a) sodium chloride (NaCl)
(b) potassium chloride (KCl)
(c) calcium chloride (CaCl2)
(d) cellulose (C6H10O5)
(e) sucrose (C12H22O11)

SOLUTION

(a) The boiling point elevation of a water solution is given by

DTb = Kbm

where Kb = 0.52&degC/m and m is the molality of the solution (the number of moles of solute particles per 1000 g of solvent). Sodium chloride is soluble in aqueous solution and yields two solute particles (Na+ and Cl-) per molecule. Thus, the molality of solute particles is (2 moles) / (1 kg) = 2 moles/kg = 2 m and

DTb = (0.52&degC/m)(2 m) = 1.04&degC

The boiling point is

Tb = 100&degC + 1.04&degC = 101.04&degC

(b) Potassium chloride is soluble in aqueous solution and yields two solute particles (K+ and Cl-) per molecule. Thus, the molality of solute particles is 2 m and the boiling point is Tb = 101.04&degC, as in part (a).

(c) Calcium chloride is soluble in aqueous solution and yields three solute particles (Ca2+ and 2Cl-) per molecule. Thus, the molality of solute particles is 3 m and

(e) Sucrose does not dissociate into more than one particle. Thus, the molality of solute particles is 1 m and

PROBLEM

Fishes in the Antarctic Ocean swim in water at about - 2&degC. (a) To prevent their blood from freezing, what must be the concentration (in molality) of the blood? Is this a reasonable physiological concentration? (b) In recent years scientists have discovered a special type of protein in these fishes' blood which, although present in quite low concentrations (< 0.001 m), has the ability to prevent the blood from freezing. Suggest a mechanism for its action.

SOLUTION

(a) The freezing point depression of water is

DTf = Kfm

where Kf is the molal freezing-point depression constant, which is 1.86 &degC/m for water. Thus,

m >DTf/Kf = (2&degC) / (1.86&degC/m) = 1.075 m

A list of reference ranges for common human blood tests appears at http://en.wikipedia.org/wiki/Reference_ranges_for_common_blood_tests. If we add up the mean concentrations of all electrolytes, metabolites, ions, and trace metals that are listed, we get 0.154 mole/L = 0.154 M. The corresponding molality is approximately 0.154 m. Thus, the molality needed to depress the freezing point by the required amount is substantially larger than the total molality of the above substances in normal human blood, and appears to be an unreasonable physiological concentration.

(b) The protein could attract water molecules and prevent them from bonding with each other and forming ice crystals.

PROBLEM

In the apparatus shown below, what will happen if the membrane is (a) permeable to both water and the Na+ and Cl- ions, (b) permeable to water and Na+ ions but not to Cl- ions, (c) permeable to water but not to Na+ and Cl- ions?

SOLUTION

(a) If the membrane is permeable to both water and the Na+ and Cl- ions, the Na+ and Cl- ions will diffuse through the membrane until both sides contain a 0.055 M solution of NaCl. The water level will be the same on both sides.

(b) If the membrane is permeable to both water and Na+ ions, but not to Cl- ions, the Na+ will not diffuse to the left because, if it did, there would be a net positive charge on the left and a net negative charge on the right. The osmotic pressure will be higher on the right side, causing the water level to increase on the right side and decrease on the left side.

(c) If the membrane is permeable to water but not to Na+ or Cl- ions, the osmotic pressure will be higher on the right, and water will diffuse from the left to the right, causing the water level to decrease on the left and increase on the right.

PROBLEM

Answer the following questions that relate to solubility of salts of lead and barium.
(a) A saturated solution is prepared by adding excess PbI2(s) to distilled water to form 1.0 L of solution at 25&degC. The concentration of Pb2+(aq) in the saturated solution is found to be 1.3 x 10-3 M. The chemical equation for the dissociation of PbI2(s) in water is shown below.

PbI2 <=> Pb2+(aq) + 2I -(aq)

(i) Write the equilibrium constant expression for the equation.
(ii) Calculate the molar concentration of I -(aq) in the solution.
(iii) Calculate the value of the equilibrium constant Ksp.

(b) A saturated solution is prepared by adding PbI2(s) to distilled water to form 2.0 L of solution at 25&degC. What are the molar concentrations of Pb2+ and I - in the solution? Justify your answer.

(c) Solid NaI is added to a saturated solution of PbI2 at 25&degC. Assuming that the volume of the solution does not change, does the molar concentration of Pb2+(aq) in the solution increase, decrease, or remain the same? Justify your answer.

(d) The value of Ksp for the salt BaCrO4 is 1.2 x 10-10. When a 500. mL sample of 8.2 x 10-6 M Ba(NO3)2 is added to 500. mL of 8.2 x 10-6 M Na2CrO4, no precipitate is observed.
(i) Assuming that volumes are additive, calculate the molar concentrations of Ba2+(aq) and CrO42-(aq) in the 1.00 L of solution.
(ii) Use the molar concentrations of Ba2+(aq) ions and CrO42-(aq) ions as determined above to show why a precipitate does not form. You must include a calculation as part of your answer.

SOLUTION

(a) (i) Ksp = [Pb2+][I -]2

(ii) For each Pb2+ in solution, there are two I -. Thus,

[I -] = 2[Pb2+] = (2)(1.3 x 10-3 M) = 2.6 x 10-3 M

(iii) Ksp = (1.3 x 10-3)(2.6 x 10-3)2 = 8.788 x 10-9

(b) The PbI2 will dissolve into Pb2+ and I - ions in the same proportion as in part (a) until [Pb2+][I -]2 is equal to Ksp = 8.788 x 10-9. Thus, the final values of [Pb2+] and [I -] will be the same as in part (a).

(c) When solid NaI is added to the saturated solution of PbI2, it breaks up into Na+ and I - ions. The concentration of I - ions increases. In order for the product [Pb2+][I -]2 to remain the same, the concentration of Pb2+ must decrease, and some PbI2 will precipitate.

(d) Nitrates are soluble in aqueous solution. Chromates containing alkali metal ions are also soluble in aqueous solution. Therefore, Ba(NO3)2 and Na2CrO4 are both soluble in aqueous solution.

(i) The molar concentrations of Ba2+(aq) and CrO42-(aq) in the final solution are both 4.1 x 10-6 M.

PROBLEM

Pure nitrosyl chloride (NOCl) gas is heated to 240&degC in a 1.00-L container. The NOCl decomposes according to

2NOCl(g) <=> 2NO(g) + Cl2(g)

At equilibrium the total pressure was 1.00 atm and the NOCl pressure was 0.64 atm.
(a) Calculate the partial pressures of NO and Cl2 in the system at equilibrium.
(b) Calculate the equilibrium constant KP.

PROBLEM

The decomposition of gaseous dimethyl ether at ordinary pressures is first order. Its half-life is 25.0 min at 500 &degC.

CH3OCH3(g) => CH4(g) + CO(g) + H2(g)

(a) Starting with 8.00 g, what mass remains (in grams) after 125 min and after 145 min?
(b) Calculate the time in minutes required to decrease 7.60 ng (nanograms) to 2.25 ng.
(c) What fraction of the original dimethyl ether remains after 150 min?

(c) Using (1), we get

q(150 min)/q0 = (1/2)(150 min / 25.0 min) = (1/2)6 = 1/64

PROBLEM

Iodide ion, I -, is oxidized to hypoiodite ion, IO-, by hypochlorite, ClO-, in basic solution according to

I -(aq) + ClO-(aq) => IO-(aq) + Cl-(aq)

Three initial-rate experiments were conducted; the results are shown in the following table:

Experiment

[I -] (mole/L)

[ClO-] (mole/L)

Initial Rate of Formation of IO- (mole/L s)

1

0.017

0.015

0.156

2

0.052

0.015

0.476

3

0.016

0.061

0.596

(a) Determine the order of the reaction with respect to each reactant.
(b) For the reaction, write the rate law that is consistent with the result of part (a), calculate the value of the specific rate constant, k, and specify units.

[from 2005 AP Chemistry Free-Response Question 3]

SOLUTION

(a) From experiments 1 and 2, for which the initial concentrations of ClO- are the same, we see that tripling the initial concentration of I - triples the rate, so the reaction is first order with respect to I -.

From experiments 1 and 3, for which the initial concentrations of I - are approximately the same, we see that quadrupling the initial concentration of ClO- quadruples the rate, so the reaction is first order with respect to ClO-.

(b) The rate law is of the form

rate = k[I -][ClO-]

The rate constant is

k = rate / [I -][ClO-]

From the three experiments in the table, we get k = 611.8, 610.3, and 610.7 L/mole s. The average of the three values is k = 610.9 L/mole s.
Back to Main Menu

PROBLEM

The catalyzed decomposition of hydrogen peroxide, H2O2(aq), is represented by the following equation:

2H2O2(aq) => 2H2O(l) + O2(g)

The kinetics of the decomposition reaction were studied and the analysis of the results show that it is a first-order reaction. Some of the experimental data are shown in the table below.

(a) Label the vertical axis of the graph.
(b) What are the units of the rate constant, k, for the decomposition of H2O2(aq)?
(c) On the graph, draw the line that represents the plot of the uncatalyzed first-order decomposition of 1.00 M H2O2(aq).

[from 2005 AP Chemistry Free-Response Question 3]

SOLUTION

(a) For a first-order reaction, the concentration of H2O2 at time t is

[H2O2](t) = [H2O2](0) e-kt

where [H2O2](0) is the concentration at time t = 0, and k is the rate constant. If we take the natural logarithms of both sides, we get

ln [H2O2](t) = ln [H2O2](t) - kt

A graph of ln [H2O2](t) vs. t is a straight line with a negative slope equal to - k and an intercept of ln [H2O2](0) with the vertical axis. The vertical axis should be labeled "ln [H2O2](t)".

(b) The rate law for the reaction is

rate = - (1/2) d[H2O2] / dt = k[H2O2]

Since the units of [H2O2] are M/s or moles/L s, the units of k are s-1.

(c) For the uncatalyzed decomposition of 1.00 M H2O2, the graph of ln [H2O2](t) vs. time would be a straight line with negative slope starting at the same point at t = 0 but decreasing at a slower rate.
Back to Main Menu

The solution of a salt derived from a strong acid and a strong base is neutral. The solution of a salt derived from a strong acid and a weak base is acidic. The solution of a salt derived from a weak acid and a strong base is basic. The solution of a salt derived from a weak acid and a weak base is (acidic, basic, approximately neutral) if Ka (>, b.

PROBLEM

Which of the following is true of pure water at a temperature of 40 &degC?
(a) It is neutral (i.e., neither acidic nor basic).
(b) It has a pH of 7.00.
(c) Both (a) and (b).
(d) Neither (a) nor (b).

SOLUTION

The ionization constant of water is Kw = 10-14 at 25 &degC. A solution of pure water at this temperature will have [H+] = [OH -] = 10-7 => pH = - log [H+] = 7.00. At other temperatures, the value of Kw is different and the values of [H+] and [OH -] will be different from their values at 25 &degC, but still equal to each other, since each ionization of H2O produces one H+ and one OH -. A neutral solution is a solution in which [H+] = [OH -], so at 40 &degC the water will still be neutral, but its pH will not be exactly 7.00. So only (a) is true.

The two chlorine ions (Cl -) each donate an electron pair (shown in pink and blue) and are the Lewis base. The tin(IV) chloride or stannic chloride (SnCl4) accepts two electron pairs and is the Lewis acid.

The nitrogen dihyride ion (NH2-) donates an electron pair (shown in blue) and is the Lewis base. The hydrogen ion (H+) (shown in green) from the ammonium ion (NH4+) accepts an electron pair and is the Lewis acid.

The ammonia (NH3) donates an electron pair (shown in pink) and is the Lewis base. The hydrogen ion (H+) accepts the electron pair and is the Lewis acid.

PROBLEM

Decide if each of the following substances should be classified as a Lewis acid or base:
(a) Mn2+(b) CH3NH2(c) H2NOH in the reaction H2NOH(aq) + HCl => [H3NOH]Cl(aq)
(d) SO2 in the reaction SO2(g) + BF3(g) <=> O2S-BF3(s)
(e) Zn(OH)2 in the reaction Zn(OH)2(s) + 2OH -(aq) <=> Zn(OH)42-(aq)

The two CN - each donate an electron pair, so the CN - is a Lewis base.

(d) Fe3+ has three missing valence electrons. It can accept an electron pair, so it is a Lewis acid.

PROBLEM

Equilibrium constants can be measured for the dissociation of Lewis acid-base complexes such as the dimethyl ether complex of BF3, (CH3)2O-BF3. The value of K (here KP) for the reaction

(CH3)2O-BF3(g) <=> BF3(g) + (CH3)2O(g)

is 0.17.
(a) Tell which product is the Lewis acid and which is the Lewis base.
(b) What is the F-B-F angle in BF3? In (CH3)2O-BF3?
(c) What is the hybridization of O in (CH3)2O-BF3? Of the boron atom?
(d) If you place 1.00 g of the complex in a 565 mL flask at 25&degC, what is the total pressure in the flask? What are the partial pressures of the Lewis acid, the Lewis base, and the complex?

Since the oxygen atom in (CH3)2O has two lone electron pairs which can bond with the boron atom in BF3, BF3 is the Lewis acid and (CH3)2O is the Lewis base.

(b) The boron atom in BF3 has three bonding electron pairs and no lone pairs surrounding it. Therefore the shape of BF3 is trigonal planar, and the F-B-F angle is 120&deg. The boron atom in (CH3)2O-BF3 has three bonding electron pairs and one lone pair surrounding it. Therefore the overall arrangement of the four electron pairs is tetrahedral, and the F-B-F angle is 109.5&deg.

(c) There are four electron pairs surrounding either the oxygen or the boron atom in (CH3)2O-BF3. Therefore, the hybridization is sp3 in both cases.

SOLUTION

[OH -] = Kw / [H+] = 10-14 / 10-3 = 10-11 M

(b) Sulfuric acid (H2SO4) is a strong acid which ionizes once according to
H2SO4 => H+ + HSO4-

HSO4- is a weak acid which ionizes according to

HSO4- => H+ + SO42-

with ionization constant Ka2 = 1.3 x 10-2. Let x be the number of moles per liter of H2SO4 which ionize only once to form HSO4- and let y be the number of moles per liter of H2SO4 which ionize twice to form SO42-. At equilibrium, we have

SOLUTION

(a) All alkali metal oxides and all alkaline earth metal oxides except beryllium oxide (BeO) dissolve in water to form basic solutions. Sodium oxide dissolves in water to form sodium hydroxide, a base:

Na2O(s) + H2O(l) => 2NaOH(aq)

(b) Calcium oxide dissolves in water to form calcium hydroxide, which is insoluble in aqueous solution:

CaO(s) + H2O(l) => Ca(OH)2(s)

(c) Aluminum oxide is amphoteric.

(d) Most nonmetallic oxides dissolve in water to form acidic solutions. Sulfur trioxide reacts with water to form sulfuric acid:

Sodium hydroxide (NaOH) is a strong base and HNO3 is a strong acid. There will be equal numbers of H+ and OH- ions, so the solution will be neutral.

(d) Cu(NO3)2 + 2H2O => Cu(OH)2 + 2HNO3

Cuprous hydroxide (Cu(OH)2) is a weak base and nitric acid (HNO3) is a strong acid, so the solution will be acidic.

PROBLEM

The following three mixtures have been prepared: CaO plus water, SiO2 plus water, and CO2 plus water. For each mixture, predict whether the pH is less than 7, equal to 7, or greater than 7. Justify your answer.

[from 2005 AP Chemistry Free-Response Question 5]

SOLUTION

CaO: Calcium oxide is basic. It reacts with water to form calcium hydroxide according to

(e) HCN(aq) + C2H3O2- <=> HC2H3O2(aq) + CN -(aq)

PROBLEM

Write the molecular and ionic equations for the following neutralizations and predict the pH (< 7, > 7, or ≈ 7) of a solution of the salt formed:
(a) phosphoric acid and calcium hydroxide
(b) acetic acid and magnesium hydroxide
(c) carbonic acid and potassium hydroxide

In a 0.1 N solution of carbonic acid, H2CO3, in water at 25&degC, only 0.29% ionization takes place. Calculate the ionization constant for carbonic acid.

SOLUTION

A 0.1 N solution of carbonic acid contains 0.1 equivalents of H2CO3 per liter of solution. Each mole of H2CO3 is equal to two equivalents of H2CO3. Therefore, a 0.1 N solution of carbonic acid contains 0.05 moles of H2CO3 per liter of solution, so it is a 0.05 M solution of H2CO3. Carbonic acid ionizes according to

The pH of a water solution of sodium carbonate may be (A) 1.5, (B) 3.5, (C) 5.5, (D) 8.5.

SOLUTION

Sodium carbonate, Na2CO3, is soluble in aqueous solution:

Na2CO3 => 2Na+ + CO32-

The carbonate can hydrolyze with H2O to form the hydrogen carbonate ion HCO3-,

CO32- + H2O <=> HCO3- + OH-

for which the hydrolysis constant is Kh = Kw/Ka2 = (10-14) / (4.8 x 10-11) = 2.08 x 10-4, where Ka2 = 4.8 x 10-11 is the ionization constant of the hydrogen carbonate ion. The carbonate can also react with H3O+ to form the hydrogen carbonate ion,

CO32- + H3O+ <=> HCO3- + H2O

for which the equilibrium constant is 1/Ka2 = 1/(4.8 x 10-11) = 2.083 x 1010. Whichever species, H2O or H3O+, the carbonate reacts with, the result will be more OH - and less H3O+, resulting in a pH which is greater than 7 or basic.

In general, the solution of a salt derived from a strong base and a weak acid, or a strong base and a neutral entity, is basic. The carbonate ion, CO32- is the conjugate base of a weak acid, HCO3-, and is therefore a strong base. The sodium ion, Na+, is neither acidic nor basic. The correct answer is (D).

Thus,

PROBLEM

A solution of equal concentrations of lactic acid and sodium acetate was found to have pH = 3.08. (a) What are the values of pKa and Ka of lactic acid? (b) What would the pH
be if the acid had twice the concentration of the salt?

SOLUTION

(a) Let L- and Ac- represent the lactate and acetate ions. Lactic and acetic acid ionize according to

Let x be the moles per liter of HL which ionize and let y be the moles per liter of Ac- which combine with H3O+ to form HAc. Assume that the initial concentrations of HL and NaAc are both 1 M. NaAc is soluble in aqueous solution. The equilibrium concentrations are

(b) Use the result from part (a), Ka1 = 4.004 x 10-2.
Let L- and Ac- represent the lactate and acetate ions. Lactic and acetic acid ionize according to

Let x be the moles per liter of HL which ionize and let y be the moles per liter of Ac- which combine with H3O+ to form HAc. Assume that the initial concentration of HL is 2 M and the initial concentration of NaAc is 1 M. NaAc is soluble in aqueous solution. The equilibrium concentrations are

Substitute (2) into (1).

Subtract y from both sides.

We can solve this graphically for y. Plot f(y) as a function of y for 0 < y < 1. We see that the value of y that makes f(y) zero is between 0.999 and 0.9999. We can narrow down the solution to y = 0.99951641. Using (2), we get x = 1.036719985 => x - y = 0.037203575 = [H3O+] => pH = - log [H3O+] = 1.429. See LacticAcetic070421.xls.

PROBLEM

One liter of a 2 M solution of lactic acid (CH3CHOHCO2H) is combined with one liter of a 2 M solution of sodium acetate (NaCH3CO2). (a) What is the pH of the resulting solution? (b) Repeat if the initial concentrations are 4 M instead of 2 M. (c) Repeat if the initial concentrations are 0.1 M instead of 2 M.

SOLUTION

(a) Let "L" and "Ac" represent the lactate (CH3CHOHCO2-) and acetate (CH3CO2-) ions, respectively. Then lactic and acetic acids ionize according to

After the two solutions are mixed, we have a solution containing 1 M lactic acid and 1 M sodium acetate. Sodium acetate is soluble in aqueous solution. The lactic acid will ionize to some extent and some of the resulting H3O+ will combine with Ac- to form HAc. Let x be the moles per liter of HL which ionize and let y be the moles per liter of Ac- which combine with H3O+ to form HAc. At equilibrium, we have

Substitute (2) into (1).

Subtract y from both sides.

We can solve this graphically for y. Plot f(y) as a function of y for 0 < y < 1. We see that the value of y that makes f(y) zero is between 0.73 and 0.74. We can narrow down the solution to y = 0.736043742. Using (2), we get x = 0.736093935 => x - y = 5.01931 x 10-5 = [H3O+] => pH = - log [H3O+] = 4.299. See LacticAcetic070407.xls.

(b) If the initial concentrations are 4 M, the solution to the problem is the same except that a = 2. x and y must both be between 0 and 2. Plot f(y) as a function of y for 0 < y < 2. The value of y that makes f(y) zero is between 1.45 and 1.50. We can narrow down the solution to y = 1.47211258. Using (2), we get x = 1.472162777 => x - y = 5.01964 x 10-5 = [H3O+] => pH = - log [H3O+] = 4.299. See LacticAcetic070407.2.xls.

(c) If the initial concentrations are 0.1 M, the solution to the problem is the same except that a = 0.05. x and y must both be between 0 and 0.05. Plot f(y) as a function of y for 0 < y < 0.05. The value of y that makes f(y) zero is between 0.0365 and 0.0370. We can narrow down the solution to y = 0.036778392. Using (2), we get x = 0.036828462 => x - y = 5.00704 x 10-5 = [H3O+] => pH = - log [H3O+] = 4.300. See LacticAcetic070407.3.xls.

HSO4- + H2O => H3O+ + SO42-

The solubility product for PbSO4 is Ksp = 1.8 x 10-8. Be sure to justify all approximations.

SOLUTION

The hydrogen sulfate or bisulfate ion, HSO4-, is a weak acid with ionization constant Ka = 1.2 x 10-2. The addition of PbSO4 to the H3O+ solution causes the H3O+ to combine with SO42-. Let x be the amount of PbSO4 which dissolves per liter of solution and let y be the amount of HSO4- produced per liter. At equilibrium, we have

[Pb2+] = x
[SO42-] = x - y
[H3O+] = 0.1 - y
[HSO4-] = y

The solubility product for PbSO4 is

Ksp = [Pb2+][SO42-] = x(x - y) (1)

The ionization constant of hydrogen sulfate is

Ka = [H3O+][SO42-] / [HSO4-] = (0.1 - y)(x - y) / y (2)

(1) and (2) are two nonlinear equations in two unknowns x and y. We can assume a value for y, solve (1) for x, substitute the result into (2) and solve for a new value of y, and keep repeating this until the values of x and y obtained converge. From (1), we get

since x cannot be negative. From (2) we get

For a trial value of y, we note that y cannot be greater than 0.1, the initial amount of H3O+ per liter of solution. So start with y = 0.05.

Substituting y = 0.05 into (3), we get x = 5 x 10-2. Substituting this into (4), we get y = (0.121, 4.15 x 10-2) from the (+, -) solution, so we reject the + solution.

We substitute y = 4.15 x 10-2 into (3) and get x = 4.15 x 10-2. We substitute this into (4) and get y = 3.50 x 10-2. If we repeat this process a number of times, the results converge to x = 4.10 x 10-4 and y = 3.66 x 10-4. Thus, the solubility of PbSO4 is 4.10 x 10-4 M.

Note: In order for this method to work, we must choose a way to solve for x and y iteratively in such a way that the solutions converge.

PROBLEM

A solution is prepared by dissolving 0.200 mole of sodium formate, HCO2Na, and 0.250 moles of formic acid, HCO2H, in approximately 200(&plusmn50) mL of water. Calculate the concentrations of H3O+ and OH-. The dissociation constant of formic acid is Ka = 1.8 x 10-4.

The ionization constant

PROBLEM

One liter of a 2 M solution of lactic acid (CH3CHOHCO2H) is combined with one liter of a 2 M solution of sodium acetate (NaCH3CO2). (a) What is the pH of the resulting solution? (b) Repeat if the initial concentrations are 4 M instead of 2 M. (c) Repeat if the initial concentrations are 0.1 M instead of 2 M.

SOLUTION

(a) Let "L" and "Ac" represent the lactate (CH3CHOHCO2-) and acetate (CH3CO2-) ions, respectively. Then lactic and acetic acids ionize according to

After the two solutions are mixed, we have a solution containing 1 M lactic acid and 1 M sodium acetate. Sodium acetate is soluble in aqueous solution. The lactic acid will ionize to some extent and some of the resulting H3O+ will combine with Ac- to form HAc. Let x be the moles per liter of HL which ionize and let y be the moles per liter of Ac- which combine with H3O+ to form HAc. At equilibrium, we have

Substitute (2) into (1).

Subtract y from both sides.

We can solve this graphically for y. Plot f(y) as a function of y for 0 < y < 1. We see that the value of y that makes f(y) zero is between 0.73 and 0.74. We can narrow down the solution to y = 0.736043742. Using (2), we get x = 0.736093935 => x - y = 5.01931 x 10-5 = [H3O+] => pH = - log [H3O+] = 4.299. See LacticAcetic070407.xls.

(b) If the initial concentrations are 4 M, the solution to the problem is the same except that a = 2. x and y must both be between 0 and 2. Plot f(y) as a function of y for 0 < y < 2. The value of y that makes f(y) zero is between 1.45 and 1.50. We can narrow down the solution to y = 1.47211258. Using (2), we get x = 1.472162777 => x - y = 5.01964 x 10-5 = [H3O+] => pH = - log [H3O+] = 4.299. See LacticAcetic070407.2.xls.

(c) If the initial concentrations are 0.1 M, the solution to the problem is the same except that a = 0.05. x and y must both be between 0 and 0.05. Plot f(y) as a function of y for 0 < y < 0.05. The value of y that makes f(y) zero is between 0.0365 and 0.0370. We can narrow down the solution to y = 0.036778392. Using (2), we get x = 0.036828462 => x - y = 5.00704 x 10-5 = [H3O+] => pH = - log [H3O+] = 4.300. See LacticAcetic070407.3.xls.

PROBLEM

A solution is prepared by adding 2.05 g of sodium acetate, NaCH3COO, to 100 mL of 0.100 M HCl solution. What is the resulting concentration of H3O+? A subsequent addition of 6.00 mL of 0.100 M HCl is made. What is the new concentration of H3O+?

SOLUTION

The initial concentration of NaCH3COO is (2.499 x 10-2 mole) / (0.100 L) = 0.2499 M. Since NaCH3COO is an ionic compound containing an alkali metal ion (Na+), it is soluble in aqueous solution, and the initial concentrations of Na+ and
CH3COO- are both 0.2499 M. Since HCl is a strong acid, the initial concentrations of H3O+ and Cl- are both 0.100 M. Acetic acid, CH3COOH, is a weak acid with ionization constant Ka = 1.85 x 10-5 (Mahan 1966, p. 193), which ionizes in water according to

CH3COOH + H2O <=> H3O+ + CH3COO-

The addition of the sodium acetate causes this reaction to proceed to the left. Let x be the number of moles per liter of H3O+ which combine with CH3COO- to form CH3COOH. At equilibrium,

The additional HCl causes more of the CH3COO- to combine with H3O+ to form CH3COOH. Let x be the number of moles per liter of CH3COO- which combine with H3O+. At equilibrium, the new concentrations are

On the (left, right) side we have a (zinc, copper) electrode immersed in a solution of (ZnSO4, CuSO4). Both ZnSO4 and CuSO4 are soluble in aqueous solution. The standard reduction potentials (25&degC and 1 M concentrations of Zn2+ and Cu2+) for the reduction of Zn2+ and Cu2+ are

Zn2+ + 2e- => Zn, E0 = - 0.76 V
Cu2+ + 2e- => Cu, E0 = + 0.34 V

Cu2+ is easier to reduce than Zn2+ because its standard reduction potential is more positive, so in the above electrochemical cell the Cu2+ will be reduced and the Zn will be oxidized. The actual half reactions that occur are

Zn => Zn2+ + 2e- => Zn, E0 = + 0.76 V
Cu2+ + 2e- => Cu, E0 = + 0.34 V

A voltmeter would measure an electromotive force (emf) of

Ecell0 = + 0.76 V + 0.34 V = + 1.10 V

The (zinc, copper) electrode is the (anode, cathode). Electrons flow through the wire from the zinc electrode to the copper electrode, from the anode to the cathode. The (anode, cathode) is the (-, +) terminal. A battery is often represented schematically by the following symbol

where the (shorter, longer) line represents the (anode, cathode). In physics, the charge carriers are thought of as positive charges which flow into the - terminal and out of the + terminal, even though physically the charge carriers are electrons which are negatively charged and flow in the opposite direction.

SOLUTION

(a) Write down the oxidation numbers of each element.

Carbon is oxidized from +3 to +4. Manganese is reduced from +7 to +2. The oxidation-reduction half reactions are

C2O42- => CO2 (oxidation)
MnO4- => Mn2+ (reduction)

We balance these two half reactions by first balancing the non-oxygen non-hydrogen elements, then balancing oxygen by adding H2O, then balancing hydrogen by adding H+, then balancing charge by adding e-. The balanced half reactions are

Add three times the first equation to two times the second equation.

3Fe + 2NO3- + 8H+ => 3Fe2+ + 2NO + 4H2O, E0 = 1.40 V

PROBLEM

The following are some substances that contain the element nitrogen: KNO3, KNO2, NH4Cl, N2H5, N2H4, NO2, N2, NH3, N2O4, HNO3.
(a) What is the oxidation state of nitrogen in each substance?
(b) In relation to nitrogen, which substances can act as oxidizing agents?
(c) In relation to nitrogen, which substances can act as reducing agents?
(d) In which of the substances is nitrogen completely oxidized?
(e) In which substance does nitrogen have the lowest oxidation number?
(f) Name the substances that can, in relation to nitrogen, act both as an oxidizing agent and as a reducing agent.

SOLUTION

(a) KNO3 (potassium nitrate):

+1+5-2
&nbsp&nbspKNO3

KNO2 (potassium nitrite):

+1+3-2
&nbsp&nbspKNO2

NH4Cl (ammonium chloride):

+5-1-1
&nbspNH4Cl

N2H5 (dinitrogen pentahydride):

+5/2-1
&nbsp&nbspN2H5

N2H4 (dinitrogen tetrahydride):

+2-1
N2H4

NO2 (nitrogen dioxide):

+4-2
&nbspNO2

N2 (nitrogen):

0
N2

NH3 (ammonia):

+3-1
&nbspNH3

N2O4 (dinitrogen tetraoxide):

+4-2
N2O4

HNO3 (nitric acid):

+1+5-2
&nbsp&nbspHNO3

(b) Since the electronic structure of nitrogen is 1s22s22p3, it can acquire up to three additional valence shell electrons, resulting in an oxidation state of -3. Thus, all of the above substances can be reduced and all can act as oxidizing agents.

(c) Nitrogen can lose up to five valence shell electrons, resulting in an oxidation state of +5. All of the substances can be oxidized and act as reducing agents except those which have oxidation states of +5: KNO3, NH4Cl, HNO3.

(d) Nitrogen is completely oxidized in KNO3, NH4Cl, and HNO3.

(e) Nitrogen has the lowest oxidation number in N2.

(f) All except KNO3, NH4Cl, and HNO3 can act both as an oxidizing agent and as a reducing agent.

or

Hg2+(aq) + Hg(l) => Hg22+, DE0 = +0.131 V, 1e- transferred

The equilibrium constant K of this reaction is given by

PROBLEM

A half-cell (A) consisting of a strip of nickel dipping into a 1 M solution of Ni2+, and a half-cell (B) consisting of a strip of zinc dipping into a 1 M solution of Zn2+ were successively connected with a standard hydrogen half-cell. The magnitudes of the individual half-cell potentials were then determined as

(a) When both the half-cells (A) and (B) were connected with the hydrogen half-cell, the metallic electrode (Ni or Zn) was found to be negative. What is the correct sign of the electrode potentials?
(b) Of the substances Ni, Ni2+, Zn, Zn2+; which is the strongest oxidant? Which is the strongest reductant?
(c) Will a noticeable reaction occur when metallic nickel is placed in a 1 M solution of Zn2+? Will a noticeable reaction occur when metallic zinc is placed in a 1 M solution of Ni2+?
(d) Zinc forms a complex ion with hydroxide ion, Zn(OH)42-. If hydroxide ion were added to half-cell (B), would its electrode potential become more positive, less positive, or be unaffected?
(e) If half-cells (A) and (B) were connected together, which electrode would be negative? What would the cell voltage be?

SOLUTION

(a) Since either metallic electrode is found to be negative, and the hydrogen electrode positive, the metallic electrode is being oxidized and is the anode, while hydrogen ions are being reduced at the hydrogen electrode, and the hydrogen electrode is the cathode. Nickel and zinc are both stronger reducing agents than hydrogen. The correct signs of the electrode potentials shown for reactions (A) and (B) are both negative.

(b) Only Ni2+ and Zn2+ are oxidants, with the following electrode potentials:

Since E0 is larger for (A) than for (B), Ni2+ is the stronger oxidant.

Only Ni and Zn are reductants, with the following electrode potentials:

(C) Ni => Ni2+ + 2e-, E0 = 0.25 V
(D) Zn => Zn2+ + 2e-, E0 = 0.77 V

Since E0 is larger for (D) than for (C), Zn is the stronger reductant.

(c) Zinc is easier to oxidize than Ni, so if metallic Ni is placed in a 1 M solution of Zn2+, no noticeable reaction will occur, but if metallic Zn is placed in a 1 M solution of Ni2+, a noticeable reaction will occur.

(d) If the hydroxide ion were added to half cell (B), Zn(OH)42- would be formed, consuming Zn2+. The zinc electrode potential would become more negative or less positive.

(e) If half-cells (A) and (B) were connected, Zn would dissolve and Ni would be deposited. Electrons would flow from the Zn electrode to the Ni electrode. The Zn electrode would be negative and the Ni electrode would be positive. The two half-cell reactions would be

Zn => Zn2+ + 2e-, E0 = 0.77 V
Ni2+ + 2e- => Ni, E0 = - 0.25 V

The cell voltage would be

DE0 = 0.77 V - 0.25 V = 0.52 V

PROBLEM

A galvanic cell consists of a strip of cobalt metal, Co, dipping into 1 M Co2+ solution, and another half-cell in which a piece of platinum dips into a 1 M solution of Cl-. Chlorine gas at 1 atm pressure is bubbled into this solution. The observed cell voltage is 1.63 V, and as the cell operates the cobalt electrode is negative. Given only that the standard potential for the chlorine-chloride ion half-cell is

(1/2)Cl2 + e- => Cl-, E0 = 1.36 V (1)

supply the desired information.
(a) What is the spontaneous cell reaction?
(b) What is the standard potential of the cobalt electrode?
(c) Would the cell voltage increase or decrease if the pressure of chlorine gas increased?
(d) What would the cell voltage be if the concentration of Co2+ were reduced to 0.01 M?

Assume that A = 1. Then from (1) we get C = 1/2. From (5) we get B = C/4 = 1/8. From (3) we get D = B = 1/8. From (4) we get E = B = 1/8. From (2) we get F = (A - D - 2E)/2 = (1 - 1/8 - 1/4)/2 = 5/16. Thus, the balanced equation is

HCl + (1/8)KMnO4 => (1/2)H2O + (1/8)KCl + (1/8)MnCl2 + (5/16)Cl2

Multiplying all coefficients by 16 we get

16HCl + 2KMnO4 => 8H2O + 2KCl + 2MnCl2 + 5Cl2

PROBLEM

Propanoic acid, HC3H5O2, ionizes in water according to

HC3H5O2(aq) <=> C3H5O2-(aq) + H+(aq), Ka = 1.34 x 10-5

(a) Write the equilibrium-constant expression for the reaction.

(b) Calculate the pH of a 0.265 M solution of propanoic acid.

(c) A 0.496 g sample of sodium propanoate, NaC3H5O2, is added to a 50.0 mL sample of a 0.265 M solution of propanoic acid. Assuming that no change in the volume of the solution occurs, calculate each of the following: (i) the concentration of the propanoate ion, C3H5O2-(aq), in the solution, (ii) the concentration of the H+(aq) ion in the solution.

(d) The methanoate ion, HCO2-(aq), reacts with water to form methanoic acid and hydroxide ion according to

HCO2-(aq) + H2O(l) <=> HCO2H(aq) + OH -(aq)

Given that [OH -] is 4.18 x 10-6 M in a 0.309 M solution of sodium methanoate, calculate each of the following: (i) the value of Kb for the methanoate ion, HCO2-(aq), (ii) the value of Ka for methanoic acid, HCO2H.

The amount of NaC3H5O2 is

The initial concentration of NaC3H5O2 is

[NaC3H5O2] = (5.163 x 10-3 mole) / (0.050 L) = 0.1033 M

NaC3H5O2 is soluble in aqueous solution. The additional C3H5O2- causes some C3H5O2- to recombine with H+ and form HC3H5O2. Let y be the number of moles per liter of C3H5O2- which recombine with H+. The equilibrium concentrations are

Kb = [HCO2H][OH -] / [HCO2-] = (4.18 x 10-6)2 / 0.309 = 5.655 x 10-11

The acid ionization constant is

Ka = Kw/Kb = 10-14 / (5.655 x 10-11) = 1.769 x 10-4

(e) Ka is larger for methanoic acid, so methanoic acid is stronger than propanoic acid.

PROBLEM

Answer the following questions about a pure compound that contains only carbon, hydrogen, and oxygen.

(a) A 0.7549 g sample of the compound burns in O2(g) to produce 1.9061 g of CO2(g) and 0.3370 g of H2O(g). (i) Calculate the individual masses of C, H, and O in the 0.7549 g sample. (ii) Determine the empirical formula for the compound.

(b) A 0.5246 g sample of the compound was dissolved in 10.0012 g of lauric acid, and it was determined that the freezing point of the lauric acid was lowered by 1.68&degC. The value of Kf of lauric acid is 3.90&degC/m. Assume that the compound does not dissociate in lauric acid. (i) Calculate the molality of the compound dissolved in the lauric acid. (ii) Calculate the molar mass of the compound from the information provided.

(c) Without doing any calculations, explain how to determine the molecular formula of the compound based on the answers to parts (a)(ii) and (b)(ii).

(d) Further tests indicate that a 0.10 M aqueous solution of the compound has a pH of 2.6. Identify the organic functional group that accounts for this pH.

The molar mass of the compound is

(c) The molecular formula of the compound can be obtained by determining the molar mass corresponding to the empirical formula of the compound and dividing the molar mass of the compound by this mass. The molar mass corresponding to the empirical formula is

Ka = x2 / (0.10 - x)

Since the pH of the solution is 2.6,

[H3O+] = 10-pH = 10-2.6 = 2.512 x 10-3 M = x

and the acid dissociation constant is

Ka = (2.512 x 10-3)2 / (0.10 - 2.512 x 10-3) = 6.472 x 10-5

PROBLEM

A solution of equal concentrations of lactic acid and sodium acetate was found to have pH = 3.08. (a) What are the values of pKa and Ka of lactic acid? (b) What would the pH
be if the acid had twice the concentration of the salt?

SOLUTION

(a) Let L- and Ac- represent the lactate and acetate ions. Lactic and acetic acid ionize according to

Let x be the moles per liter of HL which ionize and let y be the moles per liter of Ac- which combine with H3O+ to form HAc. Assume that the initial concentrations of HL and NaAc are both 1 M. NaAc is soluble in aqueous solution. The equilibrium concentrations are

(b) Use the result from part (a), Ka1 = 4.004 x 10-2.
Let L- and Ac- represent the lactate and acetate ions. Lactic and acetic acid ionize according to

Let x be the moles per liter of HL which ionize and let y be the moles per liter of Ac- which combine with H3O+ to form HAc. Assume that the initial concentration of HL is 2 M and the initial concentration of NaAc is 1 M. NaAc is soluble in aqueous solution. The equilibrium concentrations are

Substitute (2) into (1).

Subtract y from both sides.

We can solve this graphically for y. Plot f(y) as a function of y for 0 < y < 1. We see that the value of y that makes f(y) zero is between 0.999 and 0.9999. We can narrow down the solution to y = 0.99951641. Using (2), we get x = 1.036719985 => x - y = 0.037203575 = [H3O+] => pH = - log [H3O+] = 1.429. See LacticAcetic070421.xls.

PROBLEM

One liter of a 2 M solution of lactic acid (CH3CHOHCO2H) is combined with one liter of a 2 M solution of sodium acetate (NaCH3CO2). (a) What is the pH of the resulting solution? (b) Repeat if the initial concentrations are 4 M instead of 2 M. (c) Repeat if the initial concentrations are 0.1 M instead of 2 M.

SOLUTION

(a) Let "L" and "Ac" represent the lactate (CH3CHOHCO2-) and acetate (CH3CO2-) ions, respectively. Then lactic and acetic acids ionize according to

After the two solutions are mixed, we have a solution containing 1 M lactic acid and 1 M sodium acetate. Sodium acetate is soluble in aqueous solution. The lactic acid will ionize to some extent and some of the resulting H3O+ will combine with Ac- to form HAc. Let x be the moles per liter of HL which ionize and let y be the moles per liter of Ac- which combine with H3O+ to form HAc. At equilibrium, we have

Substitute (2) into (1).

Subtract y from both sides.

We can solve this graphically for y. Plot f(y) as a function of y for 0 < y < 1. We see that the value of y that makes f(y) zero is between 0.73 and 0.74. We can narrow down the solution to y = 0.736043742. Using (2), we get x = 0.736093935 => x - y = 5.01931 x 10-5 = [H3O+] => pH = - log [H3O+] = 4.299. See LacticAcetic070407.xls.

(b) If the initial concentrations are 4 M, the solution to the problem is the same except that a = 2. x and y must both be between 0 and 2. Plot f(y) as a function of y for 0 < y < 2. The value of y that makes f(y) zero is between 1.45 and 1.50. We can narrow down the solution to y = 1.47211258. Using (2), we get x = 1.472162777 => x - y = 5.01964 x 10-5 = [H3O+] => pH = - log [H3O+] = 4.299. See LacticAcetic070407.2.xls.

(c) If the initial concentrations are 0.1 M, the solution to the problem is the same except that a = 0.05. x and y must both be between 0 and 0.05. Plot f(y) as a function of y for 0 < y < 0.05. The value of y that makes f(y) zero is between 0.0365 and 0.0370. We can narrow down the solution to y = 0.036778392. Using (2), we get x = 0.036828462 => x - y = 5.00704 x 10-5 = [H3O+] => pH = - log [H3O+] = 4.300. See LacticAcetic070407.3.xls.

PROBLEM

Vitamin C is the simple compound C6H8O6. Besides being an acid, it is also a reducing agent. One method for determining the amount of vitamin C in a sample therefore is to titrate it with a solution of bromine, Br2, a good oxidizing agent:

C6H8O6(aq) + Br2(aq) => 2HBr(aq) + C6H6O6(aq)

Suppose a 1.00 g "chewable" vitamin C tablet requires 27.85 mL of 0.102 M Br2 for titration to the equivalence point. How many grams of vitamin C are in the tablet?

PROBLEM

The lead content of a sample can be estimated by converting the lead to PbO2 and then dissolving the PbO2 in an acid solution of KI. This liberates I2 according to the equation

PbO2(s) + 4H+(aq) + 2I -(aq) => Pb2+(aq) + I2(aq) + 2H2O(l) (1)

The liberated I2 is then titrated with Na2S2O3 according to

I2(aq) + 2S2O32-(aq) => 2I -(aq) + S4O62-(aq) (2)

The amount of titrated I2 is related to the amount of lead in the sample. If 0.576 g of a lead-containing mineral requires 35.23 mL of 0.0500 M Na2S2O3 for titration of the liberated I2, what is the weight percent of lead in the mineral?

SOLUTION

The amount of Na2S2O3 used for the titration is (35.23 x 10-3 L)(0.0500 M) = 1.762 x 10-3 mole of Na2S2O3. Since each mole of Na2S2O3 contains one mole of S2O32-, 1.762 x 10-3 mole of S2O32- was used. From (2), this implies that (1/2)(1.762 x 10-3 mole) = 8.808 x 10-4 mole of I2 was consumed, and from (1) this means that 8.808 x 10-4 mole of PbO2 was consumed. Each mole of PbO2 contains one mole of Pb, so 8.808 x 10-4 mole of Pb was consumed. Lead has a molar mass of 207.2 g/mole, so the amount of lead consumed was (8.808 x 10-4 mole)(207.2 g/mole) = 0.1825 g. Thus, the weight percent of lead in the mineral is (0.1825 g) / (0.576 g) = 0.3168 = 31.68%.

PROBLEM

You have done a laboratory experiment in which you obtained a new iron-containing compound. Its formula could be one of two possibilities: K3[Fe(C2O4)3] or K[Fe(C2O4)2(H2O)2]. In each of these, the C2O42- ion is the oxalate ion. To find which is correct, you dissolve 1.356 g of the compound in acid, which converts the oxalate ion to oxalic acid according to

2H+(aq) + C2O42-(aq) => H2C2O4(aq) (1)

and then titrate the oxalic acid with potassium permanganate. The balanced, net ionic equation for the titration is

The titration requires 34.50 mL of 0.108 M KMnO4. What is the correct formula of the iron-containing compound?

SOLUTION

The required number of moles of potassium permanganate (KMnO4) is (34.50 x 10-3 L)(0.108 M) = 3.726 x 10-3 mole. Potassium permanganate is soluble in aqueous solution and breaks up into K+ and MnO4- ions, so this yields 3.726 x 10-3 mole of MnO4-. According to equation (2), for every 2 moles of MnO4- consumed, 5 moles of H2C2O4 are consumed, so the number of moles of H2C2O4 used up is
(5/2)(3.726 x 10-3 mole) = 9.315 x 10-3 mole. Thus, 9.315 x 10-3 mole of C2O42- is used up.

If the unknown compound is K3[Fe(C2O4)3], the number of moles present would have been (1/3)(9.315 x 10-3 mole) = 3.105 x 10-3 mole because each mole of K3[Fe(C2O4)3] contains 3 moles of C2O42-. Since the molar mass of K3[Fe(C2O4)3] is (3)(39.0983 g/mole) + 55.847 g/mole + (3)[(2)(12.011 g/mole) + (4)(15.9994 g/mole)] = 437.2 g/mole, the mass of the sample would have been (437.2 g/mole)(3.105 x 10-3 mole) = 1.358 g.

If the unknown compound is K[Fe(C2O4)2(H2O)2], the number of moles present would have been (1/2)(9.315 x 10-3 mole) = 4.658 x 10-3 mole because each mole of K[Fe(C2O4)2(H2O)2] contains 2 moles of C2O42-. Since the molar mass of K[Fe(C2O4)2(H2O)2] is 39.0983 g/mole + 55.847 g/mole + (2)[(2)(12.011 g/mole) + (4)(15.9994 g/mole)] + (2)[(2)(1.0079 g/mole) + 15.9994 g/mole] = 307.0 g/mole, the mass of the sample would have been (307.0 g/mole)(4.658 x 10-3 mole) = 1.430 g.

The mass inferred assuming that the unknown compound is K3[Fe(C2O4)3] matches the mass of the sample. Therefore, the unknown compound is K3[Fe(C2O4)3].

A 0.10 M NaOH solution is slowly added to 25 mL of a 0.10 M HCl solution. Calculate the pH of the mixture after (a) 10 mL, (b) 25 mL, and (c) 35 mL have been added.

SOLUTION

(a) The total amount of hydrochloric acid (HCl) is (0.10 mole/L)(25 x 10-3 L) = 0.0025 mole, which is the amount of H+ since HCl is a strong acid. The total amount of sodium hydroxide (NaOH) is (0.10 mole/L)(10 x 10-3 L) = 0.0010 mole, which is the amount of OH - since NaOH is a strong base. The 0.0010 mole of OH - will neutralize 0.0010 mole of H+, leaving an excess of 0.0025 mole - 0.0010 mole = 0.0015 mole of H+. Since the total volume of the solution is 25 mL + 10 mL = 35 mL = 35 x 10-3 L, the final concentration of H+ is

[H+] = (0.0015 mole) / (35 x 10-3 L) = 4.286 x 10-2 M

and the pH is

pH = - log [H+] = - log(4.286 x 10-2) = 1.368.

(b) The total amount of hydrochloric acid (HCl) is (0.10 mole/L)(25 x 10-3 L) = 0.0025 mole, which is the amount of H+ since HCl is a strong acid. The total amount of sodium hydroxide (NaOH) is (0.10 mole/L)(25 x 10-3 L) = 0.0025 mole, which is the amount of OH - since NaOH is a strong base. The 0.0025 mole of OH - will neutralize 0.0025 mole of H+, resulting in a neutral solution with pH = 7. Since HCl is a strong acid and NaOH is a strong base, the salt formed (sodium chloride, NaCl) does not undergo hydrolysis.

(c) The total amount of hydrochloric acid (HCl) is (0.10 mole/L)(25 x 10-3 L) = 0.0025 mole, which is the amount of H+ since HCl is a strong acid. The total amount of sodium hydroxide (NaOH) is (0.10 mole/L)(35 x 10-3 L) = 0.0035 mole, which is the amount of OH - since NaOH is a strong base. The 0.0025 mole of H+ will neutralize 0.0025 mole of OH -, leaving an excess of 0.0035 mole - 0.0025 mole = 0.0010 mole of OH -. Since the total volume of the solution is 25 mL + 35 mL = 60 mL = 60 x 10-3 L, the final concentration of OH - is

A solution is made by mixing exactly 500 mL of 0.167 M NaOH with exactly 500 mL of 0.100 M CH3COOH. Calculate the equilibrium concentrations of H+, CH3COOH, CH3COO-, OH -, and Na+.

SOLUTION

Sodium hydroxide (NaOH) is a strong base which is fully ionized into Na+ and OH - ions. Acetic acid (CH3COOH) is a weak acid with ionization constant Ka = 1.8 x 10-5. After mixing, the solution has a total volume of 1 L and is 0.0835 M NaOH and 0.0500 M CH3COOH with a total of 0.0835 mole NaOH and 0.0500 mole CH3COOH. The 0.0500 mole of CH3COOH is neutralized by 0.0500 mole of NaOH, leaving 0.0335 mole of OH - and 0.0500 mole of CH3COO-. Let x be the number of moles per liter of CH3COO- which hydrolyze according to

PROBLEM

Calculate the pH at the equivalence point for the following titration: 0.10 M HCOOH versus 0.10 M NaOH.

SOLUTION

The equation for the neutralization reaction between formic acid (HCOOH), a weak acid, and sodium hydroxide (NaOH), a strong base, is

HCOOH(aq) + NaOH(aq) <=> NaCOOH(aq) + H2O, Ka = 1.7 x 10-4

Suppose we combine equal volumes of 0.1 M HCOOH and 0.1 M NaOH. The molarities of HCOOH and NaOH in the combined volume of solution are both 0.05 M. There are sufficient amounts of H+ from HCOOH and OH- from NaOH to cause neutralization, but since HCOOH is a weak acid, the formate ion (HCOO-) undergoes hydrolysis according to

Thus, [OH -] = x = 1.715 x 10-6 M and

PROBLEM

A piece of magnesium metal weighing 1.56 g is placed in 100.0 mL of 0.100 M AgNO3 at 25&degC. Calculate [Mg2+] and [Ag+] in solution at equilibrium. What is the mass of the magnesium left? The volume remains constant.

Two moles of electrons are transferred for each mole of Mg oxidized. The standard emf of the cell is related to the equilibrium constant of the redox reaction according to

From the equation for the net reaction, we see that for every two moles of Ag+ reduced, one mole of Mg is oxidized. The total number of moles of Ag is (0.1 M)(0.1 L) = 0.01 mole. The total number of moles of Mg is (1.56 g) / (24.31 g/mole) = 0.06417 mole. Since K >> 1, the reaction continues until essentially all of the Ag+ is used up. At equilibrium, the amount of Mg2+ in solution is half the initial amount of Ag+ in solution, which is (0.5)(0.01 mole) = 0.005 mole => [Mg2+] = (0.005 mole) / (0.1 L) = 0.05 M.

The equilibrium constant (Kf) for this reaction is related to the standard emf according to

PROBLEM

When 25.0 mL of a solution containing both Fe2+ and Fe3+ ions is titrated with 23.0 mL of 0.0200 M KMnO4 (in dilute sulfuric acid), all of the Fe2+ ions are oxidized to Fe3+ ions. Next, the solution is treated with Zn metal to convert all of the Fe3+ ions to Fe2+ ions. Finally, 40.0 mL of the same KMnO4 solution are added to the solution in order to oxidize the Fe2+ ions to Fe3+. Calculate the molar concentrations of Fe2+ and Fe3+ in the original solution.

The equilibrium constant is also related to the concentrations of H+, Pb2+, and H2 according to

PROBLEM

Discuss the Haber process for synthesizing ammonia. In particular:
(a) Who developed the Haber process?
(b) What is the balanced equation for the formation of ammonia from nitrogen and hydrogen?
(c) Is the reaction endothermic or exothermic?
(d) How does the catalyst that is used in the Haber process work?

SOLUTION

(a) Fritz Haber, a German chemist, discovered the process in 1905.

(b) N2(g) + 3H2(g) => 2NH3(g)

(c) The reaction is exothermic, with DH0 = - 92.6 kJ.

(d) Haber discovered that a catalyst consisting of iron plus a few percent of potassium oxides and aluminum oxides catalyze the formation of ammonia from nitrogen and hydrogen. N2 and H2 dissociate into atoms on the surface of the metal catalyst. The nitrogen and hydrogen atoms migrate on the surface and form NH3 molecules which leave the surface.

(a) Determine the value of the standard enthalpy change, DHrxn0, for the combustion of CO(g) at 298 K using the following information:

(b) Determine the value of the standard entropy change, DSrxn0, for the combustion of CO(g) at 298 K using the information in the following table:

Substance

S2980 (J/mole K)

CO(g)

197.7

CO2(g)

213.7

O2(g)

205.1

(c) Determine the standard free energy change, DGrxn0, for the reaction at 298 K.
(d) Is the reaction spontaneous under standard conditions at 298 K? Justify your answer.
(e) Calculate the value of the equilibrium constant, Keq, for the reaction at 298 K.

PROBLEM

Temperature (&degC)

Water Vapor Pressure (mm Hg)

0

4.58

20

17.54

40

55.32

60

149.38

80

355.1

(a) From the data in the above table, find the standard enthalpy of vaporization of water.
(b) Find the boiling point temperature of water when the ambient pressure is 1 atm.
(c) What is the standard entropy of vaporization of water?
(d) What is the standard free energy of vaporization of water?

SOLUTION

(a) Use the Clausius-Clapeyron equation,

ln P = - DHvap/RT + C

where P is the vapor pressure, DHvap is the standard enthalpy of vaporization, R = 8.314 J/mole K is the gas constant, T is the absolute temperature, and C is a constant. Write the above equation for any two pressure-temperature pairs, (P1, T1) and (P2, T2).

The boiling point temperature is the temperature at which the vapor pressure equals the ambient pressure, which is 1 atm = 760 mm Hg. Using the last data point in the above table for (P1, T1) and P2 = 760 mm Hg, we get

The three types of nuclear decay described are (n is a neutrino, n* is an antineutrino):

(a) The average mass number of a Cu atom is 63.55. This indicates that Cu(68, 29) has too many neutrons and will undergo beta - decay.

(b) The average mass number of a Cd atom is 112.4. This indicates that Cd(103, 48) has too few neutrons and will undergo beta + (positron) decay.

(c) Bk (atomic number Z = 97) is a transuranic (Z > 92) element. The mean atomic mass of Bk is 247. The transuranic elements tend to undergo alpha decay except for the more massive isotopes, which undergo beta - decay. Since Bk(243, 97) is one of the lighter isotopes, it undergoes alpha decay.

(d) Db (Z = 105) is also transuranic. The most stable isotope is Db(268, 105). Since Db(260, 105) is one of the lighter isotopes, it undergoes alpha decay.

PROBLEM

Within the halogen group:
(a) Which element is the most chemically active?
(b) Which element has the smallest ionic radius?
(c) Which element has the lowest first ionization potential?
(d) Which element exhibits metallic properties at room temperature?

PROBLEM

Identify the functional group that characterizes each of the following organic compounds and also give the general form of the compound:
(a) alcohol
(b) aldehyde
(c) acid
(d) ester
(e) ether
(f) ketone

SOLUTION

(a) An alcohol contains the hydroxyl (OH) functional group. A hydrocarbon where an end hydrogen is replaced by a hydroxyl group is a primary alcohol. A hydrocarbon where a non-end hydrogen is replaced by a hydroxyl group is a secondary alcohol. Two general forms of alcohols are

R-OH

and

&nbsp&nbspOH
&nbsp&nbsp|
R-C-R'
&nbsp&nbsp|
&nbsp&nbspH

where R and R' are hydrocarbon radicals.

(b) An aldehyde is formed by dehydrating an oxidized alcohol. It contains a formyl group (CHO),

PROBLEM

(a) Write the structures and names of two primary alcohols having the molecular formula C7H16O.
(b) Write the structures and names of two secondary alcohols having the molecular formula C7H16O.
(c) Write the structures and names of two tertiary alcohols having the molecular formula C7H16O.

SOLUTION

(a) A (primary, secondary, tertiary) alcohol is an alcohol in which the carbon bonded to the hydroxide is bonded to (0 or 1, 2, 3) other carbons.

It has the name "1-heptanol" because it is formed from heptane by replacing a hydrogen attached to an end carbon (a "1" carbon) with a hydroxyl group. Carbons are numbered beginning from the end closest to the hydroxyl group.

A second example of a primary alcohol with the molecular formula C7H16O is

PROBLEM

For each of the following esters, identify the alcohol and organic acid from which the ester is formed, draw the structural formula of the alcohol and organic acid, and draw the structural formula of the ester:
(a) propyl benzoate (balsam)
(b) 3-methylbutyl acetate (bananas)
(c) octyl acetate (oranges)
(d) methyl butyrate or methyl butanoate (apples)

salicylic acid + acetic acid => acetylsalicylic acid + water

Certain atomic nuclei have magnetic dipole moments due to the angular momenta of their constituent protons and neutrons. The nuclei act like bar magnets.

In the presence of an external magnetic field, these nuclei can have different alignments relative to the field direction. The different alignments are quantized (i.e., only certain alignments are possible) and associated with different energy levels.

If such a nucleus is subjected to electromagnetic radiation of the right frequency, it can absorb the radiation and make a nuclear spin transition from one of these energy levels to another. The difference between the energy levels depends on the strength of the external field, the nuclear species, and the neighboring atoms if the nucleus is part of a molecule.

By subjecting a sample of molecules to an external magnetic field of constant strength and electromagnetic radiation of varying frequency, and noting the frequencies at which the sample absorbs radiation, we can deduce the presence of certain nuclei and some of the structural characteristics of the molecule. The technique is called nuclear magnetic resonance (NMR) spectroscopy. For the magnetic field strengths typically used, resonance occurs at radio frequencies (n ~ 30 MHz - 3 GHz).

NMR spectrometers work by either keeping the magnetic field strength fixed and varying the radio frequency or by keeping the radio frequency fixed and varying the magnetic field strength.

High-resolution NMR spectroscopy can separate closely spaced resonances that appear to be a single resonance using low-resolution NMR spectrometers.

The corresponding wavelength is

where n is the frequency, h = 6.626 x 10-34 J s is Planck's constant, and c = 3 x 108 m/s is the speed of light. This wavelength is at the violet end of the visible part of the electromagnetic spectrum.

PROBLEM

With a 300 MHz NMR instrument, the three signals for 2,2-dimethyl-1-propanol are recorded at 261, 861, and 957 Hz, respectively, downfield from (CH3)4Si. Calculate the chemical shift d for each of the signals and the frequencies that would be recorded at 90 MHz.

SOLUTION

The chemical shift, in parts per million (ppm), is given by

We use this formula to calculate the chemical shift for the three signals measured with a 300 MHz instrument. Then we rewrite the formula in the form

[distance of peak from (CH3)4Si in Hz] = d[spectrometer frequency in MHz]

to calculate the position of the peaks that would be obtained with a 90 MHz NMR instrument.

300 MHz Peak

d

90 MHz Peak

261 Hz

(261 Hz) / (300 MHz) = 0.87 x 10-6 = 0.87 ppm

(0.87 x 10-6)(90 MHz) = 78.3 Hz

861 Hz

(861 Hz) / (300 MHz) = 2.87 x 10-6 = 2.87 ppm

(2.87 x 10-6)(90 MHz) = 258.3 Hz

957 Hz

(957 Hz) / (300 MHz) = 3.19 x 10-6 = 3.19 ppm

(3.19 x 10-6)(90 MHz) = 287.1 Hz

The three signals can be identified with three types of equivalent hydrogens: (1) the nine hydrogens which are bonded to the carbons in the three methyl groups, (2) the two hydrogens which are bonded to the carbon which is adjacent to the oxygen, and (3) the single hydrogen which is bonded directly to the oxygen. The type (1, 3) hydrogen has the (least, most) deshielding due to the oxygen and will have its peak at the (highest, lowest) resonant frequency. We therefore identify type (1, 2, 3) with (957, 861, 261) Hz. The relative intensities are (9, 2, 1).

&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp|&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp|&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp|&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp|&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp|&nbsp&nbsp&nbsp&nbsp&nbspNMR spectrum of 2,2-dimethyl-1-propanol
&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp|&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp|&nbsp&nbsp&nbsp&nbsp&nbsp|&nbsp&nbsp|&nbsp&nbsp|&nbsp&nbsp|&nbsp&nbsp|--+--+--+-->
frequency (Hz)
Back to Main Menu

PROBLEM

Explain the assignment of 1H NMR signals of chloro(methoxy)methane (ClCH2OCH3)in the 90 MHz NMR spectrum shown below, where the frequencies are measured relative to 90 MHz. The blue line at ~ 490 Hz is identified with the hydrogens near the chlorine; the red line at ~ 330 Hz is identified with the hydrogens in the CH3.

SOLUTION

The hydrogens near the chlorine are near two electronegative atoms, the chlorine and the oxygen. They are more deshielded from the external magnetic field than the hydrogens in the CH3. The blue hydrogens experience a stronger external field. Hence the energy difference between spin states is greater for the blue hydrogens, and these hydrogens require the higher radio frequency to achieve resonance.

If we imagine dividing the molecule in half, by a vertical plane which is perpendicular to the line joining the two center carbons, we see that both halves are equivalent. All of the methyl (CH3) groups bonded to one of the central carbons are equivalent. All of the hydrogens in a given methyl group are equivalent. Thus, all hydrogens in the molecule are equivalent, and we would expect to see one peak in the NMR spectrum.

There are three groups of equivalent hydrogens: (1) the hydrogens in the end methyl groups, (2) the four innermost hydrogens, whose carbon is bonded to the central oxygen, and (3) the four hydrogens whose carbons are bonded to the outer oxygens. We would expect to see three peaks in the NMR spectrum.

We can estimate the relative positions of the three peaks by the amount of deshielding the corresponding hydrogens experience from the oxygens in the molecule. All three groups of hydrogens are deshielded by the oxygen attached to their carbon. The differences in deshielding are determined by the more distant oxygens. The hydrogens in group (1, 2, 3) are separated from the more distant oxygens by an average of (6.5, 4.0, 4.5) bonds. The hydrogens in group (1, 2) have the (least, most), experience the (largest, smallest) net magnetic field, and require the (highest, lowest) radio frequency to achieve resonance.

It has symmetry about a vertical plane through the top carbon and the line joining the two bottom carbons. It also has symmetry about the plane of the page. All four hydrogens are equivalent, and we would see one peak in an NMR spectrum.

There are three types of equivalent hydrogens: (1) the two hydrogens that are bonded to the same carbons as the chlorines, (2) the hydrogen, on the third carbon, which is on the same side of the plane of the page as the chlorines, and (3) the hydrogen, on the third carbon, which is on the opposite side of the plane of the page relative to the chlorines. The type (1, 3) hydrogen has the (most, least) deshielding and would require the (highest, lowest) radio frequency to achieve resonance. The relative intensity of the line due to type (1, 2, 3) hydrogen is (2, 1, 1).

There are two types of equivalent hydrogens: (1) the two hydrogens that are bonded to the same carbons as the chlorines and (2) the two hydrogens bonded to the third carbon. The type (1, 2) hydrogen has the (most, least) deshielding and would require the (highest, lowest) radio frequency to achieve resonance. The relative intensities of the lines due to both types of hydrogens are the same.

Ethoxyethane (diethyl ether) has two sets of equivalent hydrogens: (1) the six hydrogens bonded to the end carbons and (2) the four hydrogens bonded to the carbons next to the oxygen. The type (1, 2) hydrogens have (less, more) deshielding by the oxygen and therefore achieve resonance at a (lower, higher) radio frequency and have a (lower, higher) chemical shift. The relative intensity of the type (1, 2) absorption is (60, 40), based on the number of hydrogens which contribute.

The hydrogens bonded to a primary carbon of an alkyl group have a chemical shift of d = 0.8-1.0 ppm (Vollhardt and Schore 1999, Table 10-2, p. 397 [VS99]). Since there is some deshielding from the oxygen, we expect that the type 1 hydrogens
will have a chemical shift near the high end of this range, d ~ 1.0 ppm. The hydrogens next to the oxygen in an ether have a chemical shift of d = 3.3-3.9 ppm (VS99), so we would expect a chemical shift of d ~ 3.6 ppm for the type 2 hydrogens.

Because of spin-spin coupling between the type 1 and type 2 hydrogens, the type 1 peak is split into a triplet with relative intensities (15, 30, 15), and the type 2 peak is split into a quartet with relative intensities (5, 15, 15, 5). For spin-spin coupling between an adjacent CH3 and CH2, which is vicinal coupling involving hydrogens from two adjacent carbons, we expect a coupling constant J ~ 6-8 Hz.

1,3-dibromopropane has two sets of equivalent hydrogens: (1) the four hydrogens bonded to the end carbons and (2) the two hydrogens bonded to the center carbon. The type (1, 2) hydrogens have (more, less) deshielding by the bromines and therefore achieve resonance at a (higher, lower) radio frequency and have a (higher, lower) chemical shift. The relative intensity of the type (1, 2) peak is (32, 16), based on the number of hydrogens which contribute. The above structure is equivalent to

When a bromine and two hydrogens are bonded to a primary carbon of a bromoalkane, the chemical shift of the hydrogens is d ~ 3.4-3.6 ppm (VS99), so the chemical shift of the type 1 hydrogens is expected to be d ~ 3.5 ppm. Hydrogens bonded to a secondary carbon of an alkyl group have a chemical shift of d ~ 1.2-1.4 ppm (VS99). We expect that the chemical shift of the type 2 hydrogens will be intermediate between the 3.4-3.6 ppm and 1.2-1.4 ppm ranges, so we estimate that d ~ 2.4 ppm. This is because the bromines are not bonded to the same carbon as these hydrogens, but to the adjacent carbons.

Because of spin-spin coupling between the type 1 and type 2 hydrogens, the type 1 peak is split into a triplet with relative intensities (8, 16, 8), and the type 2 peak is split into a quintet with relative intensities (1, 4, 6, 4, 1). For spin-spin coupling between hydrogens bonded to adjacent carbons, which is vicinal coupling, we expect a coupling constant J ~ 6-8 Hz.

2-methyl-2-butanol has four sets of equivalent hydrogens: (1) the six hydrogens that belong to the methyl groups that are bonded to the same carbon as the hydroxyl group, (2) the three hydrogens bonded to carbon at the right end of the main carbon chain, (3) the two hydrogens bonded to the number 3 carbon (recall that we number the carbons beginning at the end of the main chain closest to the hydroxyl group), and (4) the hydrogen in the hydroxyl group. The above structure is equivalent to

The type 4 hydrogen is closest to the oxygen and experiences the most deshielding. The hydrogen in an alcoholic hydroxyl group has a chemical shift of d ~ 0.5-5.0 ppm (VS99). We therefore expect that the type 4 hydrogen will have a chemical shift of d ~ 2.75 ppm, which is the mean of the range.

The type 1 and type 2 hydrogens are bonded to primary carbons and should have a chemical shift d ~ 0.8-1.0 ppm (VS99). Since the type 1 hydrogens are closer to the oxygen, we estimate that for them d ~ 1.0 ppm and for the type 2 hydrogens d ~ 0.9 ppm.

The type 3 hydrogens are bonded to a secondary carbon and should have a chemical shift of d ~ 1.2-1.4 ppm (VS99). Because of the presence of the oxygen, we estimate d ~ 1.4 ppm.

Based on the chemical shifts, we expect that the type 4 hydrogen achieves resonance at the highest frequency, the type 3 at the second highest, the type 1 at the third highest, and the type 2 at the lowest.

The relative intensity of the type (1, 2, 3, 4) peak is (60, 30, 20, 10), based on the number of hydrogens which contribute.

Because of spin-spin coupling between the type 2 and type 3 hydrogens, the type 2 peak is split into a triplet with relative intensities (7.5, 15, 7.5), and the type 3 peak is split into a quartet with relative intensities (2.5, 7.5, 7.5, 2.5). For spin-spin coupling between hydrogens bonded to adjacent carbons, which is vicinal coupling, we expect a coupling constant J ~ 6-8 Hz.

1,2,2-trichloroethane has two types of hydrogens: (1) the two hydrogens bonded to the carbon with the single chlorine and (2) the single hydrogen bonded to the carbon with two chlorines. The type (1, 2) hydrogen has (less, more) deshielding by the chlorines and therefore achieves resonance at a (lower, higher) radio frequency and has a (lower, higher) chemical shift. The relative intensity of the type (1, 2) peak is (20, 10), based on the number of hydrogens which contribute. The above structure is equivalent to

When two hydrogens and a chlorine are bonded to an end carbon of a chloroalkane, the chemical shift of each hydrogen is d ~ 3.6-3.8 ppm (VS99). We therefore expect that the type 1 hydrogens have a chemical shift of d ~ 3.7 ppm. The chemical shift of each hydrogen in a dichloromethane molecule is d = 5.3 ppm (Vollhardt and Schore 1999, p. 398). We therefore expect that the chemical shift of the type 2 hydrogen is d ~ 5.3 ppm.

Because of spin-spin coupling between the type 1 and type 2 hydrogens, the type 1 peak is split into a doublet with relative intensities (10, 10), and the type 2 peak is split into a triplet with relative intensities (2.5, 5, 2.5). For spin-spin coupling between hydrogens bonded to adjacent carbons, which is vicinal coupling, we expect a coupling constant J ~ 6-8 Hz.